Question

The harmonic conjugate of $(4,1)$ with respect to the points $(3,2)$ and$( - 1,6)$ is

Answer 1

Hint: Approach the solution by applying the section formula for given points. Here is the section formula for x coordinate and the section formula for y coordinates is similar as x coordinate.
Section formula $x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}$

Here we have to find harmonic conjugate of $(4,1)$ with respect to given points
Let $(4,1)$ divides $(3,2)$ and $( - 1,6)$ in $K:1$ ratio
So here let us apply the section formula
Section formula $x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}$
$\Rightarrow 4 = \dfrac{{k( - 1) + 1(3)}}{{k + 1}} \\\ \Rightarrow 4k + 4 = 3 - k \\\ \Rightarrow 5k = - 1 \\\ \Rightarrow k = \dfrac{{ - 1}}{5} \\\$
So, here the given points $(3,2)$and $( - 1,6)$ are going divide in $- 1:5$ ratio
Here the ratio $- 1:5$ divides the points externally but we have to divide the ratio internally
So to get the internal point ratio we have to remove the negative sign from the external ratio.
$\therefore$ Internal ratio =$1:5$
The harmonic conjugate divides the given point internally in ratio $1:5$
Apply the section formula
$x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}$
$\Rightarrow \dfrac{{1( - 1) + 5(3)}}{{5 + 1}} \\\ \Rightarrow \dfrac{7}{3} \\\$
$y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}} \\\ \Rightarrow y = \dfrac{{1(6) + 5(2)}}{{5 + 1}} \\\ \Rightarrow y = \dfrac{8}{3} \\\$
Therefore the harmonic conjugate of the required point that divides internally in the ratio $1:5$ = $\left( {\dfrac{8}{3},\dfrac{7}{3}} \right)$

Note: In these types of problems external or internal ratio matter where sign value is different. Here we have used section formulas to both x and y coordinates.